POJ-1195-Mobile phones（二维树状数组）

Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The squares form an S * S matrix with the rows and columns numbered from 0 to S-1. Each square contains a base station. The number of active mobile phones inside a square can change because a phone is moved from a square to another or a phone is switched on or off. At times, each base station reports the change in the number of active phones to the main base station along with the row and the column of the matrix. 

Write a program, which receives these reports and answers queries about the current total number of active mobile phones in any rectangle-shaped area. 

Input

The input is read from standard input as integers and the answers to the queries are written to standard output as integers. The input is encoded as follows. Each input comes on a separate line, and consists of one instruction integer and a number of parameter integers according to the following table. 

The values will always be in range, so there is no need to check them. In particular, if A is negative, it can be assumed that it will not reduce the square value below zero. The indexing starts at 0, e.g. for a table of size 4 * 4, we have 0 <= X <= 3 and 0 <= Y <= 3. 

Table size: 1 * 1 <= S * S <= 1024 * 1024 
Cell value V at any time: 0 <= V <= 32767 
Update amount: -32768 <= A <= 32767 
No of instructions in input: 3 <= U <= 60002 
Maximum number of phones in the whole table: M= 2^30 

Output

Your program should not answer anything to lines with an instruction other than 2. If the instruction is 2, then your program is expected to answer the query by writing the answer as a single line containing a single integer to standard output.

Sample Input

0 4
1 1 2 3
2 0 0 2 2 
1 1 1 2
1 1 2 -1
2 1 1 2 3 
3

Sample Output

3 
4

题解：

题意就是给你4种操作：

0 S 表示输入矩阵的边长为S

1 X Y K 表示在（X，Y）的位置加K

2 L B R T 表示要你求出（L，B）~（R，T）的和

3 表示结束程序

还有

表格大小：1 * 1 <= S * S <= 1024 * 1024
任何时候的单元格值V：0 <= V <= 32767
更新金额：-32768 <= A <= 32767
输入的指令数：3 <= U <= 60002
整个桌面最大电话数：M = 2 ^ 30

这里我们想到用二维树状数组来做，具体怎么做就是分成两维，每次x和y都像一维一样+或-lowbit

不过有一个坑就是他给你的是矩阵上的点，你树状数组的时候要坐标要+1

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define ll long long
#define N 1030
using namespace std;
int T,n,x,y,k,l,b,r,t;
ll tr[N][N];
int lowbit(int x){ return x&(-x); }
void add(int x,int y,int k){
    for (int i=x;i<=n;i+=lowbit(i))
        for (int j=y;j<=n;j+=lowbit(j))
            tr[i][j]+=k;
}
ll query(int l,int b){
    ll s=0;
    for (int i=l;i>=1;i-=lowbit(i))
        for (int j=b;j>=1;j-=lowbit(j))
            s+=tr[i][j];
    return s;
}
int main(){
    while (~scanf("%d",&T)){
        if (T==3) break;
        if (T==0) scanf("%d",&n); else
        if (T==1){
            scanf("%d%d%d",&x,&y,&k);
            add(x+1,y+1,k);
        } else 
        if (T==2){
            scanf("%d%d%d%d",&l,&b,&r,&t);
            l++; b++; r++; t++;
            printf("%lld\n",query(r,t)+query(l-1,b-1)-query(r,b-1)-query(l-1,t));
        }
    }
    return 0;
}